I have no idea what any of that means. Oh well.
–
anonFeb 27 '12 at 3:40

1

If $T(n+1)=T(n)+a_n$ then $T(n+1)=a_n+a_{n-1}+\cdots+a_2+a_1+T(1)$ by applying the recurrence $n$ times. (I incorrectly said $T(n)$ instead of $T(n+1)$ in my comment above.) My assumption is that $E$ is the operator sending the function $n\to T(n)$ to $n\to T(n+1)-T(n)$, in which case it is not true that $(E-1)^3T$ is identically zero, as point 4 states.
–
anonFeb 27 '12 at 3:55

1

I don’t see a way to avoid induction completely, but in my answer I suggest a way to make it a very easy, obvious induction. (Don’t worry about the mistranslation; you included enough information to make clear what you meant.)
–
Brian M. ScottFeb 27 '12 at 4:05

2 Answers
2

Define a new function $S(n)=2^{T(n)}$. Then $S(1)=2^a$, and for $n\ge 1$ we have

$$S(n+1) = 2^{T(n) + \log_2(n)}=nS(n)\;.$$

From this it’s clear that $S(2)=1\cdot 2^a$, $S(3)=2\cdot1\cdot2^a$, and in general $S(n)=(n-1)!2^a$. Thus, $$T(n)=\log_2S(n)=\log_2 (n-1)!2^a=a+\log_2(n-1)!\;,$$ so your question boils down to asking whether $a+\log_2(n-1)!\,$ is $O(n^2)$.

Clearly $\log_2n!\le\log_2n^n=n\log_2n$, and you should easily be able to tell whether $n\log_2n$ is $O(n^2)$.

But in any case, you're making the problem too hard on yourself. You aren't being asked to solve the recursion for a closed form -- you're just being asked to show it's $O(n^2)$. In other words, you're being asked if there is a $C$ such that $T(n) < C n^2$, and $T$ lends itself nicely to an inductive argument:

$T(1) < C$

Assuming that $T(n) < C x^2$, can you prove $T(n+1) < C (x+1)^2$

While you could solve the recurrence, as the other answer has shown, it's a lot more work than you actually need to do to answer the question at hand.